You'll need to multiply B by a 3-by-3 matrix (being the product of square matrices E and F). I think you can use the 2-by-2 you've provided as part of the 3-by-3 version of F that you need.

F = [ 0 1 0 ] [ 1 0 0 ] [ 0 0 1 ]

FB = [-1 2 ] [ 9 1 ] [ 0 10 ]

Now you need to find a 3-by-3 matrix E so that EFB equals A. Since the last row of A contains 3 and 4, you need a matrix which is provide the row-operation(s) necessary to create this from what you've already got (if I'm understanding the process correctly).

stapel_eliz wrote:You'll need to multiply B by a 3-by-3 matrix (being the product of square matrices E and F). I think you can use the 2-by-2 you've provided as part of the 3-by-3 version of F that you need.

F = [ 0 1 0 ] [ 1 0 0 ] [ 0 0 1 ]

FB = [-1 2 ] [ 9 1 ] [ 0 10 ]

Now you need to find a 3-by-3 matrix E so that EFB equals A. Since the last row of A contains 3 and 4, you need a matrix which is provide the row-operation(s) necessary to create this from what you've already got (if I'm understanding the process correctly).

E(FB) = [ 1 0 0 ][-1 2 ] [ 0 1 0 ][ 9 1 ] [ a b 1 ][ 0 10 ]

When you multiply this out, you should get a system of equations:

-a + 9b = 32a + 3b = 10

Solve for the values of "a" and "b", and thus for E.

Check my work!

that is not the system you will get...

I think it is quite obvious that we need

since if a,b are both non zero we don't have an elementary matrix...at least according to any definition I have ever seen.